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\author{学号 \underline{\hspace{4cm}} \hspace{1cm} 姓名 \underline{\hspace{4cm}} }
\title{实变函数测验 5C 解答}
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\date{2024 年 5 月 27 日}
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\begin{document}

\maketitle

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圈出叙述错误的步骤，并加以改正。

\begin{enumerate}

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\item  %Problem 01
设 $f(x) = \frac{\sin(1/x)}{x^\alpha},\,\, 0<x\le 1,$
讨论 $\alpha$ 为何值时，$f(x)$ 在 $(0,1]$ 上是勒贝格可积的。

%证明：
\begin{enumerate}[label={(\arabic*)}]

\item  %1
当 $\alpha\ge 1$ 时，对 $0< x\le 1$, 有 $x^\alpha \ge x$, 因此 
$$
\int_0^1 \frac{|\sin(1/x)|}{x^\alpha}dx \le \int_0^1 \frac{|\sin(1/x)|}{x}dx. 
$$

\item  %2
设变量代换 $t=\frac{1}{x}$, 则上述右边的积分化为
$$
\int_1^{\infty}t\cdot |\sin(t)|\cdot \frac{1}{t^2}dt = \int_1^{\infty} \frac{|\sin(t)|}{t}dt = \infty. 
$$

\item  %3
因此当 $\alpha\ge 1$ 时，$f(x)$ 在 $(0,1]$ 上不是勒贝格可积的。

\item  %4
当 $\alpha<1$ 时，
$$
\int_0^1 \frac{|\sin(1/x)|}{x^\alpha}dx \le \int_0^1 \frac{1}{x^\alpha}dx
=(R)\int_0^1 \frac{1}{x^\alpha}dx = \frac{1}{1-\alpha}x^{1-\alpha}\mid_0^1 =\frac{1}{1-\alpha} 
<\infty. 
$$

\item  %6
因此当 $\alpha< 1$ 时，$f(x)$ 在 $(0,1]$ 上是勒贝格可积的。

\end{enumerate}

{\color{red}解答：
(1) 当 $\alpha\ge 1$ 时，对 $0< x\le 1$, 有 $x^\alpha \le x$, 因此 
$$
\int_0^1 \frac{|\sin(1/x)|}{x^\alpha}dx \ge \int_0^1 \frac{|\sin(1/x)|}{x}dx. 
$$


}

\vspace{0.3cm}


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\item  %Problem 02
证明：
$$\lim\limits_{n\to\infty} \int_0^\infty \frac{dt}{\left( 1+\frac{t}{n} \right)^n t^{\frac{1}{n}}} =1. $$

%证明：
\begin{enumerate}[label={(\arabic*)}]

\item  %1
对每个正整数 $n$, 记 $f_n(t) = \frac{1}{\left( 1+\frac{t}{n} \right)^n t^{\frac{1}{n}}}$, 定义在 $E=(0,\infty)$ 上。

\item  %2
对每个 $t\in E$, 极限 $\lim\limits_{n\to\infty} f_n(t) = \frac{1}{e^t}$. 代入积分可得 $\int_E\frac{1}{e^t}dt =1$. 

\item  %3
因此只需要找到一个控制函数，然后应用控制收敛定理就可以了。

\item  %4
当 $0< t\le 1$ 时，设 $n\ge 2$, 则有 $|f_n(t)|\le \frac{1}{t^{1/2}}$, 而且 $\int_{(0,1]} t^{-1/2}dt 
%= \frac{1}{-\frac{1}{2}+1}t^{-1/2+1}\mid_0^1 
= 2$. 

\item  %5
当 $t\ge 1$ 时，设 $n\ge 2$, 根据下述二项式展开，有 $|f_n(t)|\le \frac{3}{t^2}$, 而且计算可得 $\int_{[1,\infty)} \frac{3}{t^2}dt=3$. 
$$\left(1+\frac{t}{n} \right)^n = 1 + t + \frac{n(n-1)}{2} \frac{t^2}{n^2}+\cdots \ge 
\frac{n-1}{2n}t^2\ge \frac{1}{3}t^2. $$

\item  %6
因此可以取函数
$$
F(t) = \left\{\begin{array}{ll}
t^{-1/2}, & 0< t\le 1, \\ 
3t^{-2}, & t>1. 
\end{array}\right. 
$$
则 $|f_n(t)|\le F(t)$, 且 $F(t)$ 是勒贝格可积的。

\end{enumerate}

{\color{red}解答：
(5) 当 $t\ge 1$ 时，设 $n\ge 3$, 根据下述二项式展开，有 $|f_n(t)|\le \frac{3}{t^2}$, 而且计算可得 $\int_{[1,\infty)} \frac{3}{t^2}dt=3$.  

}

\vspace{0.3cm}


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\item  %Problem 03
设函数 $f(x)$ 是区间 $[a,b]$ 上的有界函数，则 $f(x)$ 在 $[a,b]$ 上黎曼可积的充分必要条件是 $f(x)$ 在 $[a,b]$ 上几乎处处连续，即不连续的点的全体组成一个零测度集。

%证明：
\begin{enumerate}[label={(\arabic*)}]

\item  %1
对每个正整数 $n$, 取一个有 $P(n)$ 个小区间的划分，使得这些小区间的最大长度随 $n$ 增加而趋于零，
 $$ a = x_0^{(n)}<x_1^{(n)}<\cdots<x_{P(n)}^{(n)} = b. $$

\item  %2
在每个小区间 $[x_{i-1}^{(n)}, x_{i}^{(n)}]$ 中，函数 $f(x)$ 的上确界和下确界分别记为 $M_i^{(n)}$ 和 $m_i^{(n)}$. 

\item  %3
函数 $f(x)$ 的达布上积分和达布下积分 $\underline{I}$ 分别为 
\begin{subequations}
\begin{align}
\overline{I} &= \lim\limits_{n\to\infty} \sum\limits_{i=1}^{P(n)} M_i^{(n)}(x_{i}^{(n)} - x_{i-1}^{(n)}), \nonumber \\ 
\underline{I} &= \lim\limits_{n\to\infty} \sum\limits_{i=1}^{P(n)} m_i^{(n)}(x_{i}^{(n)} - x_{i-1}^{(n)}). \nonumber 
\end{align}
\end{subequations}

\item  %4
对每个正整数 $n$, 根据上面取定的划分，定义一个简单函数 
$$
h_n(x) = \left\{\begin{array}{ll}
M_i^{(n)} - m_i^{(n)}, & 若 x_{i-1}^{(n)}<x< x_{i}^{(n)}, \\
0, & 若 x = x_0^{(n)}, x_1^{(n)}, \cdots, x_{P(n)}^{(n)}.  
\end{array}\right. 
$$

\item  %5
对每个 $x\in [a,b]$, 定义 $f(x)$ 在点 $x$ 的振幅为 
$$\omega(x) = \lim\limits_{\delta\to 0^+} \sup \{ |f(y)-f(z)| : y,z \in (x-\delta, x+\delta) \cap [a,b] \}. $$

\item  %6
函数 $f(x)$ 在 $x_0$ 连续当且仅当 $\omega(x_0)=0$. 

\item  %7
对 $x\neq x_i^{(n)}, 1\le i \le P(n), n=1,2,\cdots$, 有 $\lim\limits_{n\to\infty} h_n(x)=\omega(x)$. 因此 $h_n(x)$ 几乎处处收敛于 $\omega(x)$. 

\item  %8
因为 $f(x)$ 是区间 $[a,b]$ 上的有界函数，所以存在实数 $M$ 使得对任意 $x\in [a,b]$ 都有 $|f(x)|\le M$. 

\item  %9
因为 $-M\le f(x)\le M$, 所以 $-M\le h_n(x)\le M$. 

\item  %10
对函数序列 $\{h_n(x)\}$ 应用勒贝格控制收敛定理，可得 
\begin{subequations}
\begin{align}
\lim\limits_{n\to\infty} (L) \int_{[a,b]} h_n(x)dx = (L)\int_{[a,b]}\omega (x)dx.  
\tag{*}
\label{main-step-1}
\end{align}
\end{subequations}

\item  %11
根据简单函数的勒贝格积分的定义，函数 $h_n(x)$ 的勒贝格积分为 
\begin{subequations}
\begin{align}
(L)\int_{[a,b]} h_n(x)dx = \sum\limits_{i=1}^{P(n)} (M_i^{(n)} - m_i^{(n)}) (x_{i}^{(n)} - x_{i-1}^{(n)}). 
\tag{**}
\label{main-step-2}
\end{align}
\end{subequations}

\item  %12
根据 (3) 中的达布上下积分的定义，从 (\ref{main-step-1}) 和 (\ref{main-step-2}) 可知，
$$
(L)\int_{[a,b]} \omega(x) dx = \overline{I} - \underline{I}. 
$$ 

\item  %13
设 $f(x)$ 在区间 $[a,b]$ 上是黎曼可积的，则达布上积分 $\overline{I}$ 和达布下积分 $\underline{I}$ 相等。于是 $\omega(x)$ 的勒贝格积分为零。但是因为 $\omega(x)$ 是非负函数，所以 $\omega(x)$ 几乎处处为零。

\item  %14
另一方面，若 $\omega(x)$ 几乎处处为零，则 $\omega(x)$ 的勒贝格积分为零。于是 $f(x)$ 的达布上积分 $\overline{I}$ 和达布下积分 $\underline{I}$ 相等，所以 $f(x)$ 在区间 $[a,b]$ 上是黎曼可积的。

\end{enumerate}

{\color{red}解答：
(9) 因为 $-M\le f(x)\le M$, 所以 $0\le h_n(x)\le 2M$. 


}

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\end{enumerate}


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